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catsrule332

  • 3 years ago

An 11g bullet is fired with a muzzle velocity of 679m/s. What is the kinetic energy of the bullet? If the bulle hits a target that is attached to a spring with a spring constant of 1500N/m how far will the spring compress when struck by the bullet?

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  1. vikash4exploring
    • 3 years ago
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    m=11g = 0.011kg, v = 679 m/s, KE = (1/2)mv^2=(1/2)*0.011*(679)^2=2535.7255 Joules PE of spring = (1/2)kx^2, here k = spring constant, x = compression distance Assuming conservation of energy and all KE of the bullet is converted to the potential energy of the spring when it hits the spring. so KE of bullet=PE of spring (1/2)mv^2=(1/2)kx^2 =>(1/2)kx^2=2535.7255 =>kx^2=5071.451 =>x^2=5071.451/1500=3.38096 => x = square root of 3.38096=1.8387 meters = 183.87 cm

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